Millikan Oil Drop Data Analysis:

Transcription

1 Millikn Oil Drop Dt Anlysis: The experiment consists o rising tiny, electriclly chrged oil drop in n electric ield nd then lowering it gin. To rise it you pply constnt electric ield on the drop tht orces it upwrd. To lower the drop you cn either turn o the electric ield nd just let it ll or you cn reverse the ield nd orce it downwrd. Becuse o ir drg orces on the oil drop, the drop very quickly reches its terminl velocity nd moves with constnt velocity. Hence the totl orce on the oil drop is exctly zero while you observe it. Generlly you rise nd lower the drop over ixed distnce, d, nd mesure the rise nd ll times, t r nd t, respectively. You need to know the electric ield s well. This is just the voltge on the cpcitor pltes, V, divided by the distnce between them, D. The rw dt in the mesurement consists o the 5 quntities, d, t r, t, V nd D. Generlly you estblish nd mesure d nd D once nd or ll t the beginning o the experiment. (D is set by the pprtus. The lst time I mesured it ws (.0 ± 0.0) mm. ) V is estblished nd recorded or ech drop nd the sme voltge (except or the sign) is used to rise nd lower the drop i you choose to orce it down. The two t s should be mesured mny times or ech drop; isn t overdoing it. Minimizing rndom errors nd keeping trck o them is one o the two most importnt ingredients o successul oil drop experiment. I you mke N mesurements o one o the t s, the best estimte o t will just be the men (or verge) o those mesurements. The uncertinty in t will be the stndrd devition o the N mesurements divided by N. You wnt to mke N lrge to get good estimted o the stndrd devition o the mesurements nd to drive the uncertinty o the men down. The other importnt ingredient o successul experiment is the selection o relly smll drops becuse they tend to hve the ewest number o chrges on them. I drops hve, or just ew chrges on them, 0% precision in the chrge mesurement will show you chrge quntiztion. To see the dierence between 00 nd 0 chrges, you hve to mesure the chrge to ew prts in 000, much hrder tsk. Below the clssicl mechnics needed to extrct the chrge on ech drop rom the rw dt is presented. Finlly, ew comments to get you strted thinking bout how to go rom the chrges to e nd chrge quntiztion re oered. The ollowing tble contins the symbols we ll use. Tble : Symbols Used in Dt Anlysis Symbol Deinition Comment d D t r t V Rise/Fll Distnce Plte spcing Rise time Fll time Voltge cross pltes (.0 ± 0.0) mm. Free-ll or orced down by ield; your choice.

2 Tble : Symbols Used in Dt Anlysis Symbol Deinition Comment Rise velocity Fll velocity d t r d t E Electric ield in cpcitor E V D ρ Mss density o the oil Mesure it. It s little less thn g/cm. σ ρ' ρ σ Mss density o ir Clculte it rom wht you know bout ir. (It s smll enough tht high precision is not importnt here.) Including σ tkes cre o buoyncy orces. η P Viscosity o ir Viscosity o ir corrected or men ree pth eects Rdius o oil drop Atmospheric pressure It is unction o the temperture o the room. See upcoming discussion. This mtters or the size drops you will use. See the discussion below. T Temperture o the room We ll use C (degrees Celsius). b Constnt used in clculting η rom η. 0 b See up coming discussion. torr-cm q The chrge on the drop Equtions o Motion o the Drops The dt nlysis is bsed on pplying F m to the rising nd lling drops. We will tke up s the positive direction. When the drop is rising in the ield m 0 qe 4 πρ' g 6πηv () r In () the let side is zero becuse the drop is moving t its constnt terminl velocity; qe is the upwrd orce on it; the next term is the drop s weight pulling it downwrd corrected or the buoyncy o the ir, nd the lst term is the viscous ir drg on the drop. The lst term is clled Stokes lw. It is the drg orce on sphere o rdius moving through luid o viscosity η. nd is pproprite or smll drops t low velocities so tht the ir low round the drop is lminr nd

3 thereore not turbulent. I the drop is moving up, the drg on it is downwrd nd thereore comes in with - sign in (). I you just let the drop ll, its eqution o motion will be 0 4 πρ' g + 6πηv () Finlly, i you reverse the polrity insted o turning it o nd let qe help grvity pull the drop down, the eqution o motion is 4 0 qe πρ' g + 6πηv () I you look t () long enough, you will discover tht ter you tke the mesurements nd igure out nd E, there re two unknowns in it, nd q. At this point there re two options depending on whether you let the drop ree ll in zero ield or orced the drop down by reversing polrity on the pltes. Solving the Equtions o Motion or q Option : Free Fll I you mesured the ll time in zero ield you cn solve () or. The result is 9ηv - (4) ρ'g Now we cn solve (). Use () in () to eliminte the mg term (second on the right) nd then substitute the right side o (4) or. The result is 6πD 9v q η ( v V ρ'g + ). (5) The D nd V in (5) come rom writing E in terms o the plte voltge nd seprtion.

4 Option : Reversing Polrity In this cse you cn dd () nd () to get 8 πρ' g 6πη( v ). (6) Notice tht will lwys be bigger thn so there cn not be sign problems. (6) is solved or ; 9η( ) 4ρ'g. (7) Now you cn subtrct () rom () to get qe 6πη( + ). (8) Then substitute (7) into (8) q πd 9v ( ) η ( v V 4ρ'g + ) (9) Two comments re worth mking. First, you must remember tht the in the ree-ll cse (Eqs. (), (4) nd (5)) is not the sme thing s in the cse where the ield is reversed to help orce the drop downwrd (Eqs. (), (6), (7), (8) nd (9)). The ll velocities when the ield helps pull the drop down must be bigger thn the ree-ll velocity. Furthermore, little thought bout the orces involved will (should) convince you tht in the ree ll cse is proportionl to the weight o the drop nd tht in the ield orced drop cse ( ) is proportionl to twice the weight nd hence is twice s big. This is why the in the rdicls in (4) nd (5) becomes 4 in (7) nd (9). Similrly, in the ree ll cse, ( + ) is proportionl to qe; in the ield ided drop, it is proportionl to qe. Tht is why the 6 in (5) becomes in (9). 4

5 There re two things tht complicte the results in (5) nd (9) nd both hve to do with the viscosity o ir, η. First, η is unction o temperture. To good ccurcy T T (0) The units o the viscosity s given in (5) re gm cm - s - lso clled Poise. In (0) T is the Celsius temperture. The constnt out in ront is obviously the viscosity t T 0 C. You must use (0) s the irst step in evluting the viscosity. Worse yet, there is nother hiding in η. Stokes lw ssumes tht the luid tht the sphere is moving through is continuous substnce. In continuous luid, luid prticles re in immedite contct with their neighbors, nd thus in some sense couple to their surroundings in mximlly eicient wy. A rel gs is not continuous medium but swrm o prticles tht must ly or distnce clled the men ree pth beore they become wre o conditions in their neighborhood. Thus rel gs is less ble to orgnize resistnce to n object moving through it thn continuous luid would be. The dierence strts to become importnt only when the moving object is smll, mening comprble with men ree pth or smller. Thus the resistnce orce on smll object moving through gs is less thn wht you would clculte bsed on mesurement o the viscosity mde with big object. is viscosity mesured with big objects, where the men ree pth doesn t mtter. Your object is going to be µm (or even less) rdius sphere nd the men ree pth in the surrounding ir hs been mesured s µm. So i you use s the viscosity you will overestimte the drg orce nd thus get q wrong. A irly serious nlysis o the sitution (M.D. Allen nd O. G. Rbe, Re-evlution o Millikn s Oil Drop Dt or Smll Prticles in Air, J. Aerosol Sci., 57 (98)) suggests tht the viscosity tht should be used in our experiment is η - b. () + - P In this eqution comes rom (0); P is the pressure o the ir in the room in torr (mmhg); nd b torr-cm. So or µm nd P 760 torr, denomintor in () corrects η by lmost 8%. (Millikn ws wre o ll this; he just used slightly lrger vlue o b.) µm rdius drops were the biggest ones I used when I did this experiment ew yers go.) 5

6 The esiest thing to do t this point is to substitute () into () in the ree ll cse or (6) in the ield orced drop cse nd solve gin or. In both cses you get qudrtic eqution tht hs only one positive root. The results re 9η 0 b 4P () ρ'g P b in the ree-ll cse, nd 9η 0 ( ) 4P + - b - () 4ρ'g P b when the ield helps push the droplet down. To wrp ll o this up, in the ree ll cse () is substituted into (5) to give q 9 6πD V ρ'g -v ( v + ). b + - P (4) nd () is used to clculte. In the ield orced ll cse () is substituted into (9) to give nd () is used to clculte. Wht else? 9 πd q V 4ρ'g - v v ( ) ( r + ). (5) b + - P The whole gme here is to determine the q on bunch o dierent drops. To mesure the chrge on the drops you wnt to hve drops with only ew chrges on them, nd the uncertinty in the vlues o q you derive must be well estblished by your dt nlysis nd smller thn the chrge quntum. I the errors re bigger thn tht, the chrge will look continuous. So you must use the propgtion o errors ormuls you presumbly got tught in P50 something to clculte the uncertinty in ech nd every q. Needless to sy, Mthemtic (or whtever similr thing you like) mkes world o dierence to the lbor involved in this. But the error nlysis relly isn t optionl. 6

7 So you ve t lst got the chrges on bunch o dierent drops. Now wht? Once you hve the chrges on bunch o drops wht do you do? You don t do wht most students do t this point. They () orget bout clculting errors in q; () divide ech chrge they mesure 9 by.60 0 C ; () throw out the rctionl prt o the result o ech division; (4) multiply 9 ech quotient by.60 0 C ; (5) notice tht ech o the resulting chrge vlues is perect integrl multiple o e, nd (6) declre victory nd perect greement with the ccepted vlue o e. There re usully enough extr steps nd other nonsense round to cmoulge this process to some extent, but this is the hert o wht they do. This is n obviously misleding wy to process the dt. You should resist the tempttion to use wht you lredy think is the right nswer in your dt nlysis. Here s n lterntive suggestion Find chrge tht when divided into ech one o your mesured chrges gives result within n error br or two o n integer or ech nd every chrge. The smllest chrge you mesure or the smllest dierence between two chrges should be good cndidte or the number to divide by. I you ve done good experiment this will work. Next you wnt to dither tht divisor round just little to minimize the sum o the squre o the dierences between the ech division result nd the integer nerest it. Mesure the distnce to ech nerest integer in units o the error in the division result. (This is minimizing χ with respect to divisor. Look bck t P50 something.) The lrgest such number you cn ind is your vlue o e. Conused? You ll hve plenty o time to think bout it while timing drops. Your instructor my lso hve even better suggestions, or mybe be ble to help you mke sense out o this one. 7

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